In the electroysis of $A g N O_{3}$ solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol $^{- 1}$ ).

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Solution

Given: Mass of Ag deposited = 0.7 g
Molar mass of Ag = 107.9 g mol $^{- 1}$
Quantity of electricity, Q = ?
The half-cell reaction is
$A g_{(a q)}^{+} + e^{-} \to A g_{(s)}$
$∵$ 1 mole of Ag produced = 107.9 g Ag requires 1 mole of electrons.
$∴$ 0.7 g Ag will require,
$= \frac{Mass of Ag}{Molar mass of Ag}$
$= \frac{0.7}{107.9} = 6.49 \times 10^{- 3} m o l o f A g$
$∵$ 1 mole electron carry 96500C charge.
$∴ 6.49 \times 10^{- 3}$ mol of electrons will carry
$Q = \frac{96500 \times 6.49 \times 10^{- 3}}{1}$
$= 626$ coulombs.

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In the electroysis of AgNO3 solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol−1).

In the electroysis of $A g N O_{3}$ solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol $^{- 1}$ ).